NECO 2023 · Paper 2 · Q4

(a)
Solve the equation $27^{2x - 1} \times \left(\frac13\right)^{-(3x + 2)} = 9^{x + 3}$ .
(b)
Differentiate $(2x^2 + 5)^4$ with respect to $x$ .

Worked solution (try it first)

(a)

Write every number as a power of 3: $27 = 3^3$ , $\frac13 = 3^{-1}$ and $9 = 3^2$ .
Then $27^{2x - 1} = 3^{6x - 3}$ , $\left(\frac13\right)^{-(3x + 2)} = 3^{3x + 2}$ and $9^{x + 3} = 3^{2x + 6}$ .
Multiplying powers of 3 adds the indices: $3^{(6x - 3) + (3x + 2)} = 3^{9x - 1}$ .
So $3^{9x - 1} = 3^{2x + 6}$ , and the indices are equal: $9x - 1 = 2x + 6$ , $7x = 7$ , $x = 1$ .

(b)

Chain rule: bring down the power, reduce it by one, then multiply by the derivative of the inside.
$\frac{d}{dx}(2x^2 + 5)^4 = 4(2x^2 + 5)^3 \times 4x$
$= 16x(2x^2 + 5)^3$ .

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